\(A=x^2-6x+2013\)
\(\Rightarrow A=x^2-6x+9+2004\)
\(\Rightarrow A=\left(x^2-6x+9\right)+2004\)
\(\Rightarrow A=\left(x^2-2.x.3+3^2\right)+2004\)
\(\Rightarrow A=\left(x-3\right)^2+2004.\)
Ta có: \(\left(x-3\right)^2\ge0\) \(\forall x.\)
\(\Rightarrow\left(x-3\right)^2+2004\ge2004\) \(\forall x.\)
\(\Rightarrow A\ge2004.\)
Dấu '' = '' xảy ra khi:
\(\left(x-3\right)^2=0\)
\(\Rightarrow x-3=0\)
\(\Rightarrow x=3\)
Vậy \(MIN_A=2004\) khi \(x=3.\)
Chúc bạn học tốt!